3.513 \(\int \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=135 \[ \frac {2 (a+b \tan (c+d x))^{5/2}}{5 b d}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}+\frac {i (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {i (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d} \]

[Out]

I*(a-I*b)^(3/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d-I*(a+I*b)^(3/2)*arctanh((a+b*tan(d*x+c))^(1/2)
/(a+I*b)^(1/2))/d-2*b*(a+b*tan(d*x+c))^(1/2)/d+2/5*(a+b*tan(d*x+c))^(5/2)/b/d

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Rubi [A]  time = 0.24, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3543, 3482, 3539, 3537, 63, 208} \[ \frac {2 (a+b \tan (c+d x))^{5/2}}{5 b d}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}+\frac {i (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {i (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(3/2),x]

[Out]

(I*(a - I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d - (I*(a + I*b)^(3/2)*ArcTanh[Sqrt[a + b*
Tan[c + d*x]]/Sqrt[a + I*b]])/d - (2*b*Sqrt[a + b*Tan[c + d*x]])/d + (2*(a + b*Tan[c + d*x])^(5/2))/(5*b*d)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

\begin {align*} \int \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} \, dx &=\frac {2 (a+b \tan (c+d x))^{5/2}}{5 b d}-\int (a+b \tan (c+d x))^{3/2} \, dx\\ &=-\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 b d}-\int \frac {a^2-b^2+2 a b \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 b d}-\frac {1}{2} (a-i b)^2 \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} (a+i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 b d}-\frac {\left (i (a-i b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}+\frac {\left (i (a+i b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 b d}+\frac {(a-i b)^2 \operatorname {Subst}\left (\int \frac {1}{-1-\frac {i a}{b}+\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {(a+i b)^2 \operatorname {Subst}\left (\int \frac {1}{-1+\frac {i a}{b}-\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}\\ &=\frac {i (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {i (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 b d}\\ \end {align*}

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Mathematica [A]  time = 1.07, size = 158, normalized size = 1.17 \[ \frac {\frac {2 (a+b \tan (c+d x))^{5/2}}{b}+5 (b+i a) \left (-\sqrt {a+b \tan (c+d x)}+\sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )\right )+5 i (a+i b) \left (\sqrt {a+b \tan (c+d x)}-\sqrt {a+i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(3/2),x]

[Out]

((2*(a + b*Tan[c + d*x])^(5/2))/b + 5*(I*a + b)*(Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]]
- Sqrt[a + b*Tan[c + d*x]]) + (5*I)*(a + I*b)*(-(Sqrt[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]
]) + Sqrt[a + b*Tan[c + d*x]]))/(5*d)

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fricas [B]  time = 1.24, size = 4304, normalized size = 31.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/20*(20*sqrt(2)*b*d^5*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 +
3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(3/4)*sqrt((9*
a^4*b^2 - 6*a^2*b^4 + b^6)/d^4)*arctan(((3*a^10 + 11*a^8*b^2 + 14*a^6*b^4 + 6*a^4*b^6 - a^2*b^8 - b^10)*d^4*sq
rt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) + (3*a^13 + 14*a^11*b^2 +
25*a^9*b^4 + 20*a^7*b^6 + 5*a^5*b^8 - 2*a^3*b^10 - a*b^12)*d^2*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) + sqrt(
2)*((3*a^4*b + 2*a^2*b^3 - b^5)*d^7*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4
+ b^6)/d^4) + 2*(3*a^7*b + 5*a^5*b^3 + a^3*b^5 - a*b^7)*d^5*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))*sqrt((a^6
+ 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2
- 6*a^2*b^4 + b^6))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)
/d^4)^(3/4) + sqrt(2)*(d^7*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^
4) + 2*(a^3 + a*b^2)*d^5*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a
^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt(((9*a^8*b
^2 + 12*a^6*b^4 - 2*a^4*b^6 - 4*a^2*b^8 + b^10)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c)
+ sqrt(2)*(2*(9*a^5*b^3 - 6*a^3*b^5 + a*b^7)*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) +
(9*a^8*b^3 + 12*a^6*b^5 - 2*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt((a
*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4) + (9*a^11*b^2 +
21*a^9*b^4 + 10*a^7*b^6 - 6*a^5*b^8 - 3*a^3*b^10 + a*b^12)*cos(d*x + c) + (9*a^10*b^3 + 21*a^8*b^5 + 10*a^6*b^
7 - 6*a^4*b^9 - 3*a^2*b^11 + b^13)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c)))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6)/d^4)^(3/4))/(9*a^14*b^2 + 39*a^12*b^4 + 61*a^10*b^6 + 35*a^8*b^8 - 5*a^6*b^10 - 11*a^4*b^12 - a^2*b^14 + b
^16))*cos(d*x + c)^2 + 20*sqrt(2)*b*d^5*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^
6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4
)^(3/4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4)*arctan(-((3*a^10 + 11*a^8*b^2 + 14*a^6*b^4 + 6*a^4*b^6 - a^2*b
^8 - b^10)*d^4*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) + (3*a^13
+ 14*a^11*b^2 + 25*a^9*b^4 + 20*a^7*b^6 + 5*a^5*b^8 - 2*a^3*b^10 - a*b^12)*d^2*sqrt((9*a^4*b^2 - 6*a^2*b^4 +
b^6)/d^4) - sqrt(2)*((3*a^4*b + 2*a^2*b^3 - b^5)*d^7*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4
*b^2 - 6*a^2*b^4 + b^6)/d^4) + 2*(3*a^7*b + 5*a^5*b^3 + a^3*b^5 - a*b^7)*d^5*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6
)/d^4))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)
/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3*a^4*b^2 +
3*a^2*b^4 + b^6)/d^4)^(3/4) - sqrt(2)*(d^7*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*
a^2*b^4 + b^6)/d^4) + 2*(a^3 + a*b^2)*d^5*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))*sqrt((a^6 + 3*a^4*b^2 + 3*a
^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6
))*sqrt(((9*a^8*b^2 + 12*a^6*b^4 - 2*a^4*b^6 - 4*a^2*b^8 + b^10)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/
d^4)*cos(d*x + c) - sqrt(2)*(2*(9*a^5*b^3 - 6*a^3*b^5 + a*b^7)*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^
4)*cos(d*x + c) + (9*a^8*b^3 + 12*a^6*b^5 - 2*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c))*sqrt((a^6 + 3*a^4*b^
2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^
4 + b^6))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4)
+ (9*a^11*b^2 + 21*a^9*b^4 + 10*a^7*b^6 - 6*a^5*b^8 - 3*a^3*b^10 + a*b^12)*cos(d*x + c) + (9*a^10*b^3 + 21*a^
8*b^5 + 10*a^6*b^7 - 6*a^4*b^9 - 3*a^2*b^11 + b^13)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c)))*((a^6 + 3*a^4*b^
2 + 3*a^2*b^4 + b^6)/d^4)^(3/4))/(9*a^14*b^2 + 39*a^12*b^4 + 61*a^10*b^6 + 35*a^8*b^8 - 5*a^6*b^10 - 11*a^4*b^
12 - a^2*b^14 + b^16))*cos(d*x + c)^2 + 5*sqrt(2)*((a^3*b - 3*a*b^3)*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6)/d^4)*cos(d*x + c)^2 - (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*d*cos(d*x + c)^2)*sqrt((a^6 + 3*a^4*b^2 + 3*a^
2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6)
)*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4)*log(((9*a^8*b^2 + 12*a^6*b^4 - 2*a^4*b^6 - 4*a^2*b^8 + b^10)
*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) + sqrt(2)*(2*(9*a^5*b^3 - 6*a^3*b^5 + a*b^7)*d
^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) + (9*a^8*b^3 + 12*a^6*b^5 - 2*a^4*b^7 - 4*a^2*b^
9 + b^11)*d*cos(d*x + c))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2
+ 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*(
(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4) + (9*a^11*b^2 + 21*a^9*b^4 + 10*a^7*b^6 - 6*a^5*b^8 - 3*a^3*b^1
0 + a*b^12)*cos(d*x + c) + (9*a^10*b^3 + 21*a^8*b^5 + 10*a^6*b^7 - 6*a^4*b^9 - 3*a^2*b^11 + b^13)*sin(d*x + c)
)/((a^2 + b^2)*cos(d*x + c))) - 5*sqrt(2)*((a^3*b - 3*a*b^3)*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)
*cos(d*x + c)^2 - (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*d*cos(d*x + c)^2)*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 +
b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*((a^6
+ 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4)*log(((9*a^8*b^2 + 12*a^6*b^4 - 2*a^4*b^6 - 4*a^2*b^8 + b^10)*d^2*sqr
t((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) - sqrt(2)*(2*(9*a^5*b^3 - 6*a^3*b^5 + a*b^7)*d^3*sqrt(
(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) + (9*a^8*b^3 + 12*a^6*b^5 - 2*a^4*b^7 - 4*a^2*b^9 + b^11
)*d*cos(d*x + c))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*
b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3
*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4) + (9*a^11*b^2 + 21*a^9*b^4 + 10*a^7*b^6 - 6*a^5*b^8 - 3*a^3*b^10 + a*b^
12)*cos(d*x + c) + (9*a^10*b^3 + 21*a^8*b^5 + 10*a^6*b^7 - 6*a^4*b^9 - 3*a^2*b^11 + b^13)*sin(d*x + c))/((a^2
+ b^2)*cos(d*x + c))) - 8*(a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8 + (a^8 - 3*a^6*b^2 - 15*a^4*b^4 - 17*a^2*b^6
- 6*b^8)*cos(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*cos(d*x + c)*sin(d*x + c))*sqrt((a*cos(d*x
+ c) + b*sin(d*x + c))/cos(d*x + c)))/((a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*d*cos(d*x + c)^2)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.27, size = 842, normalized size = 6.24 \[ \frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 b d}-\frac {2 b \sqrt {a +b \tan \left (d x +c \right )}}{d}+\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, a}{4 d b}-\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2}}{4 d b}+\frac {b \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}-\frac {2 b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, a}{4 d b}+\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2}}{4 d b}-\frac {b \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}+\frac {2 b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+b*tan(d*x+c))^(3/2),x)

[Out]

2/5*(a+b*tan(d*x+c))^(5/2)/b/d-2*b*(a+b*tan(d*x+c))^(1/2)/d+1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(
2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a-1/4/d/b*ln(b*tan
(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*
a^2+1/4/d*b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^
2)^(1/2)+2*a)^(1/2)-2/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*
a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2
)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)-1/4/d/b*ln((a+b*tan(d*x+c))^(1
/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2
)*a+1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^
2)^(1/2)+2*a)^(1/2)*a^2-1/4/d*b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^
2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+2/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1
/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*
(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^2, x)

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mupad [B]  time = 9.32, size = 1141, normalized size = 8.45 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(a + b*tan(c + d*x))^(3/2),x)

[Out]

atan((b^6*(a + b*tan(c + d*x))^(1/2)*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4*d^2))
^(1/2)*32i)/((b^8*16i)/d - (16*a*b^7)/d - (a^2*b^6*32i)/d + (32*a^3*b^5)/d - (a^4*b^4*48i)/d + (48*a^5*b^3)/d)
+ (32*a*b^5*(a + b*tan(c + d*x))^(1/2)*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4*d^
2))^(1/2))/((b^8*16i)/d - (16*a*b^7)/d - (a^2*b^6*32i)/d + (32*a^3*b^5)/d - (a^4*b^4*48i)/d + (48*a^5*b^3)/d)
- (a^2*b^4*(a + b*tan(c + d*x))^(1/2)*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4*d^2)
)^(1/2)*96i)/((b^8*16i)/d - (16*a*b^7)/d - (a^2*b^6*32i)/d + (32*a^3*b^5)/d - (a^4*b^4*48i)/d + (48*a^5*b^3)/d
) - (96*a^3*b^3*(a + b*tan(c + d*x))^(1/2)*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4
*d^2))^(1/2))/((b^8*16i)/d - (16*a*b^7)/d - (a^2*b^6*32i)/d + (32*a^3*b^5)/d - (a^4*b^4*48i)/d + (48*a^5*b^3)/
d))*((3*a*b^2 + a^2*b*3i - a^3 - b^3*1i)/(4*d^2))^(1/2)*2i + atan((b^6*(a + b*tan(c + d*x))^(1/2)*((b^3*1i)/(4
*d^2) - a^3/(4*d^2) + (3*a*b^2)/(4*d^2) - (a^2*b*3i)/(4*d^2))^(1/2)*32i)/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^
8*16i)/d + (32*a^3*b^5)/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d) - (32*a*b^5*(a + b*tan(c + d*x))^(1/2)*((b^3*1i)
/(4*d^2) - a^3/(4*d^2) + (3*a*b^2)/(4*d^2) - (a^2*b*3i)/(4*d^2))^(1/2))/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^8
*16i)/d + (32*a^3*b^5)/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d) - (a^2*b^4*(a + b*tan(c + d*x))^(1/2)*((b^3*1i)/(
4*d^2) - a^3/(4*d^2) + (3*a*b^2)/(4*d^2) - (a^2*b*3i)/(4*d^2))^(1/2)*96i)/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b
^8*16i)/d + (32*a^3*b^5)/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d) + (96*a^3*b^3*(a + b*tan(c + d*x))^(1/2)*((b^3*
1i)/(4*d^2) - a^3/(4*d^2) + (3*a*b^2)/(4*d^2) - (a^2*b*3i)/(4*d^2))^(1/2))/((a^2*b^6*32i)/d - (16*a*b^7)/d - (
b^8*16i)/d + (32*a^3*b^5)/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d))*((3*a*b^2 - a^2*b*3i - a^3 + b^3*1i)/(4*d^2))
^(1/2)*2i + ((2*a^2)/(b*d) - (2*(a^2 + b^2))/(b*d))*(a + b*tan(c + d*x))^(1/2) + (2*(a + b*tan(c + d*x))^(5/2)
)/(5*b*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+b*tan(d*x+c))**(3/2),x)

[Out]

Integral((a + b*tan(c + d*x))**(3/2)*tan(c + d*x)**2, x)

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